A distress flare is fired into the air from a ship at sea. The height, \(h\) , in metres, of the flare above sea level is modelled by the quadratic function

\[h\,(t) =  - 0.2{t^2} + 16t + 12\,,\,t \geqslant 0\,,\]

where \(t\) is the time, in seconds, and \(t = 0\,\) at the moment the flare was fired.

Write down the height from which the flare was fired.

Find the height of the flare \(15\) seconds after it was fired.

The flare fell into the sea \(k\) seconds after it was fired.

Find the value of \(k\) .

Find \(h'\,(t)\,.\)

i)     Show that the flare reached its maximum height \(40\) seconds after being fired.

ii)    Calculate the maximum height reached by the flare.

The nearest coastguard can see the flare when its height is more than \(40\) metres above sea level.

Determine the total length of time the flare can be seen by the coastguard.

\(12\,({\text{m}})\)       (A1)

\((h\,(15) = ) - 0.2 \times {15^2} + 16 \times 15 + 12\)       (M1)

Note: Award (M1) for substitution of \(15\) in expression for \(h\).

\( = 207\,({\text{m}})\)       (A1)(G2)

\(h\,(k) = 0\)       (M1)

Note: Award (M1) for setting \(h\) to zero.

\((k = )\,\,\,80.7\,({\text{s}})\,\,\,(80.7430)\)       (A1)(G2)

Note: Award at most (M1)(A0) for an answer including \(K =  - 0.743\) .
Award (A0) for an answer of \(80\) without working.

\(h'\,(t) =  - 0.4t + 16\)        (A1)(A1)

Note: Award (A1) for \( - 0.4t\), (A1) for \(16\). Award at most (A1)(A0) if extra terms seen. Do not accept \(x\) for \(t\).

i)     \( - 0.4t + 16 = 0\)       (M1)

Note: Award (M1) for setting their derivative, from part (d), to zero, provided the correct conclusion is stated and consistent with their \(h'\,(t)\).

OR

\(t = \frac{{ - 16}}{{2 \times ( - 0.2)}}\)       (M1)

Note: Award (M1) for correct substitution into axis of symmetry formula, provided the correct conclusion is stated.

\(t = \,\,40\,({\text{s}})\)       (AG)

ii)    \( - 0.2 \times {40^2} + 16 \times 40 + 12\)       (M1)

Note: Award (M1) for substitution of \(40\) in expression for \(h\).

\( = 332\,({\text{m}})\)       (A1)(G2)

\(h\,(t) = 40\)       (M1)

Note: Award (M1) for setting \(h\) to \(40\). Accept inequality sign.

OR

M1

Note: Award (M1) for correct sketch. Indication of scale is not required.

\(78.2 - 1.17\,\,(78.2099...\,\, - 1.79005...)\)       (A1)

Note: Award (A1) for \(1.79\) and \(78.2\) seen.

(total time \( = \)) \(76.4\,({\text{s}})\,\,\,(76.4198...)\)       (A1)(G2)

Note: Award (G1) if the two endpoints are given as the final answer with no working.

Question 3: Quadratic function, problem solving.
Parts (a) (finding the initial height) and (b) (finding the height after 15 seconds), were done very well by the majority of candidates. Many struggled to translate question (c) to find the (positive) zeros of the function, or did not write that down, losing a possible method mark. The derivative in part (d) was no problem for most; only very few used \(x\) instead of \(t\). The maximum height reached was calculated correctly by the majority of candidates, but many lost the mark in part (e)(i) as they simply substituted 40 into their derivative or calculated the height at points close to 40. Only a few candidates showed correct method for part (f). Several were still able to obtain 2 marks as a result of “trial and error” of integer values for \(t\). Some candidate seem to have a problem with the notation “\(h\,(t) = \,...\)”, where this is interpreted as \(h \times t\), resulting in incorrect answers throughout.

Question 3: Quadratic function, problem solving.

Parts (a) (finding the initial height) and (b) (finding the height after 15 seconds), were done very well by the majority of candidates. Many struggled to translate question (c) to find the (positive) zeros of the function, or did not write that down, losing a possible method mark. The derivative in part (d) was no problem for most; only very few used \(x\) instead of \(t\). The maximum height reached was calculated correctly by the majority of candidates, but many lost the mark in part (e)(i) as they simply substituted 40 into their derivative or calculated the height at points close to 40. Only a few candidates showed correct method for part (f). Several were still able to obtain 2 marks as a result of “trial and error” of integer values for \(t\). Some candidate seem to have a problem with the notation “\(h\,(t) = \,...\)”, where this is interpreted as \(h \times t\), resulting in incorrect answers throughout.

Question 3: Quadratic function, problem solving.

Parts (a) (finding the initial height) and (b) (finding the height after 15 seconds), were done very well by the majority of candidates. Many struggled to translate question (c) to find the (positive) zeros of the function, or did not write that down, losing a possible method mark. The derivative in part (d) was no problem for most; only very few used \(x\) instead of \(t\). The maximum height reached was calculated correctly by the majority of candidates, but many lost the mark in part (e)(i) as they simply substituted 40 into their derivative or calculated the height at points close to 40. Only a few candidates showed correct method for part (f). Several were still able to obtain 2 marks as a result of “trial and error” of integer values for \(t\). Some candidate seem to have a problem with the notation “\(h\,(t) = \,...\)”, where this is interpreted as \(h \times t\), resulting in incorrect answers throughout.

Question 3: Quadratic function, problem solving.

Parts (a) (finding the initial height) and (b) (finding the height after 15 seconds), were done very well by the majority of candidates. Many struggled to translate question (c) to find the (positive) zeros of the function, or did not write that down, losing a possible method mark. The derivative in part (d) was no problem for most; only very few used \(x\) instead of \(t\). The maximum height reached was calculated correctly by the majority of candidates, but many lost the mark in part (e)(i) as they simply substituted 40 into their derivative or calculated the height at points close to 40. Only a few candidates showed correct method for part (f). Several were still able to obtain 2 marks as a result of “trial and error” of integer values for \(t\). Some candidate seem to have a problem with the notation “\(h\,(t) = \,...\)”, where this is interpreted as \(h \times t\), resulting in incorrect answers throughout.

Question 3: Quadratic function, problem solving.

Parts (a) (finding the initial height) and (b) (finding the height after 15 seconds), were done very well by the majority of candidates. Many struggled to translate question (c) to find the (positive) zeros of the function, or did not write that down, losing a possible method mark. The derivative in part (d) was no problem for most; only very few used \(x\) instead of \(t\). The maximum height reached was calculated correctly by the majority of candidates, but many lost the mark in part (e)(i) as they simply substituted 40 into their derivative or calculated the height at points close to 40. Only a few candidates showed correct method for part (f). Several were still able to obtain 2 marks as a result of “trial and error” of integer values for \(t\). Some candidate seem to have a problem with the notation “\(h\,(t) = \,...\)”, where this is interpreted as \(h \times t\), resulting in incorrect answers throughout.

Question 3: Quadratic function, problem solving.

Parts (a) (finding the initial height) and (b) (finding the height after 15 seconds), were done very well by the majority of candidates. Many struggled to translate question (c) to find the (positive) zeros of the function, or did not write that down, losing a possible method mark. The derivative in part (d) was no problem for most; only very few used \(x\) instead of \(t\). The maximum height reached was calculated correctly by the majority of candidates, but many lost the mark in part (e)(i) as they simply substituted 40 into their derivative or calculated the height at points close to 40. Only a few candidates showed correct method for part (f). Several were still able to obtain 2 marks as a result of “trial and error” of integer values for \(t\). Some candidate seem to have a problem with the notation “\(h\,(t) = \,...\)”, where this is interpreted as \(h \times t\), resulting in incorrect answers throughout.